JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Definite Integrals

Let

f : R \to R

be a continuous function such that

f (x) + f (x + 1) = 2

, for all

x \in R

. If

I_{1} = \int_{0}^{8} f (x) d x

and

I_{2} = \int_{- 1}^{3} f (x) d x

, then the value of

I_{1} + 2 I_{2}

is equal to ____

Answer: 16

Solution

Replacing

x

x + 1

f (x) + f (x + 1) = 2

gives

f (x + 1) + f (x + 2) = 2

, so

f (x + 2) = f (x)

and

f

has period

2

. Over one period

\int_{0}^{2} f (x) d x = \int_{0}^{1} f (x) d x + \int_{0}^{1} f (x + 1) d x = \int_{0}^{1} [f (x) + f (x + 1)] d x = 2

The interval

[0, 8]

covers four periods and

[- 1, 3]

covers two, so

I_{1} = 8

and

I_{2} = 4

, and therefore

I_{1} + 2 I_{2} = 8 + 2 \cdot 4 = 16

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